3.19.43 \(\int \frac {(a+b x) (d+e x)^{3/2}}{a^2+2 a b x+b^2 x^2} \, dx\)

Optimal. Leaf size=86 \[ -\frac {2 (b d-a e)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{b^{5/2}}+\frac {2 \sqrt {d+e x} (b d-a e)}{b^2}+\frac {2 (d+e x)^{3/2}}{3 b} \]

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Rubi [A]  time = 0.04, antiderivative size = 86, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.121, Rules used = {27, 50, 63, 208} \begin {gather*} \frac {2 \sqrt {d+e x} (b d-a e)}{b^2}-\frac {2 (b d-a e)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{b^{5/2}}+\frac {2 (d+e x)^{3/2}}{3 b} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((a + b*x)*(d + e*x)^(3/2))/(a^2 + 2*a*b*x + b^2*x^2),x]

[Out]

(2*(b*d - a*e)*Sqrt[d + e*x])/b^2 + (2*(d + e*x)^(3/2))/(3*b) - (2*(b*d - a*e)^(3/2)*ArcTanh[(Sqrt[b]*Sqrt[d +
e*x])/Sqrt[b*d - a*e]])/b^(5/2)

Rule 27

Int[(u_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[u*Cancel[(b/2 + c*x)^(2*p)/c^p], x] /; Fr
eeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

Rule 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]

Rubi steps

\begin {align*} \int \frac {(a+b x) (d+e x)^{3/2}}{a^2+2 a b x+b^2 x^2} \, dx &=\int \frac {(d+e x)^{3/2}}{a+b x} \, dx\\ &=\frac {2 (d+e x)^{3/2}}{3 b}+\frac {(b d-a e) \int \frac {\sqrt {d+e x}}{a+b x} \, dx}{b}\\ &=\frac {2 (b d-a e) \sqrt {d+e x}}{b^2}+\frac {2 (d+e x)^{3/2}}{3 b}+\frac {(b d-a e)^2 \int \frac {1}{(a+b x) \sqrt {d+e x}} \, dx}{b^2}\\ &=\frac {2 (b d-a e) \sqrt {d+e x}}{b^2}+\frac {2 (d+e x)^{3/2}}{3 b}+\frac {\left (2 (b d-a e)^2\right ) \operatorname {Subst}\left (\int \frac {1}{a-\frac {b d}{e}+\frac {b x^2}{e}} \, dx,x,\sqrt {d+e x}\right )}{b^2 e}\\ &=\frac {2 (b d-a e) \sqrt {d+e x}}{b^2}+\frac {2 (d+e x)^{3/2}}{3 b}-\frac {2 (b d-a e)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{b^{5/2}}\\ \end {align*}

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Mathematica [A]  time = 0.06, size = 77, normalized size = 0.90 \begin {gather*} \frac {2 \sqrt {d+e x} (-3 a e+4 b d+b e x)}{3 b^2}-\frac {2 (b d-a e)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{b^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((a + b*x)*(d + e*x)^(3/2))/(a^2 + 2*a*b*x + b^2*x^2),x]

[Out]

(2*Sqrt[d + e*x]*(4*b*d - 3*a*e + b*e*x))/(3*b^2) - (2*(b*d - a*e)^(3/2)*ArcTanh[(Sqrt[b]*Sqrt[d + e*x])/Sqrt[
b*d - a*e]])/b^(5/2)

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IntegrateAlgebraic [A]  time = 0.13, size = 90, normalized size = 1.05 \begin {gather*} \frac {2 \sqrt {d+e x} (-3 a e+b (d+e x)+3 b d)}{3 b^2}-\frac {2 (a e-b d)^{3/2} \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x} \sqrt {a e-b d}}{b d-a e}\right )}{b^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[((a + b*x)*(d + e*x)^(3/2))/(a^2 + 2*a*b*x + b^2*x^2),x]

[Out]

(2*Sqrt[d + e*x]*(3*b*d - 3*a*e + b*(d + e*x)))/(3*b^2) - (2*(-(b*d) + a*e)^(3/2)*ArcTan[(Sqrt[b]*Sqrt[-(b*d)
+ a*e]*Sqrt[d + e*x])/(b*d - a*e)])/b^(5/2)

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fricas [A]  time = 0.47, size = 188, normalized size = 2.19 \begin {gather*} \left [-\frac {3 \, {\left (b d - a e\right )} \sqrt {\frac {b d - a e}{b}} \log \left (\frac {b e x + 2 \, b d - a e + 2 \, \sqrt {e x + d} b \sqrt {\frac {b d - a e}{b}}}{b x + a}\right ) - 2 \, {\left (b e x + 4 \, b d - 3 \, a e\right )} \sqrt {e x + d}}{3 \, b^{2}}, -\frac {2 \, {\left (3 \, {\left (b d - a e\right )} \sqrt {-\frac {b d - a e}{b}} \arctan \left (-\frac {\sqrt {e x + d} b \sqrt {-\frac {b d - a e}{b}}}{b d - a e}\right ) - {\left (b e x + 4 \, b d - 3 \, a e\right )} \sqrt {e x + d}\right )}}{3 \, b^{2}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)*(e*x+d)^(3/2)/(b^2*x^2+2*a*b*x+a^2),x, algorithm="fricas")

[Out]

[-1/3*(3*(b*d - a*e)*sqrt((b*d - a*e)/b)*log((b*e*x + 2*b*d - a*e + 2*sqrt(e*x + d)*b*sqrt((b*d - a*e)/b))/(b*
x + a)) - 2*(b*e*x + 4*b*d - 3*a*e)*sqrt(e*x + d))/b^2, -2/3*(3*(b*d - a*e)*sqrt(-(b*d - a*e)/b)*arctan(-sqrt(
e*x + d)*b*sqrt(-(b*d - a*e)/b)/(b*d - a*e)) - (b*e*x + 4*b*d - 3*a*e)*sqrt(e*x + d))/b^2]

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giac [A]  time = 0.19, size = 112, normalized size = 1.30 \begin {gather*} \frac {2 \, {\left (b^{2} d^{2} - 2 \, a b d e + a^{2} e^{2}\right )} \arctan \left (\frac {\sqrt {x e + d} b}{\sqrt {-b^{2} d + a b e}}\right )}{\sqrt {-b^{2} d + a b e} b^{2}} + \frac {2 \, {\left ({\left (x e + d\right )}^{\frac {3}{2}} b^{2} + 3 \, \sqrt {x e + d} b^{2} d - 3 \, \sqrt {x e + d} a b e\right )}}{3 \, b^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)*(e*x+d)^(3/2)/(b^2*x^2+2*a*b*x+a^2),x, algorithm="giac")

[Out]

2*(b^2*d^2 - 2*a*b*d*e + a^2*e^2)*arctan(sqrt(x*e + d)*b/sqrt(-b^2*d + a*b*e))/(sqrt(-b^2*d + a*b*e)*b^2) + 2/
3*((x*e + d)^(3/2)*b^2 + 3*sqrt(x*e + d)*b^2*d - 3*sqrt(x*e + d)*a*b*e)/b^3

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maple [B]  time = 0.05, size = 167, normalized size = 1.94 \begin {gather*} \frac {2 a^{2} e^{2} \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )}{\sqrt {\left (a e -b d \right ) b}\, b^{2}}-\frac {4 a d e \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )}{\sqrt {\left (a e -b d \right ) b}\, b}+\frac {2 d^{2} \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )}{\sqrt {\left (a e -b d \right ) b}}-\frac {2 \sqrt {e x +d}\, a e}{b^{2}}+\frac {2 \sqrt {e x +d}\, d}{b}+\frac {2 \left (e x +d \right )^{\frac {3}{2}}}{3 b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x+a)*(e*x+d)^(3/2)/(b^2*x^2+2*a*b*x+a^2),x)

[Out]

2/3*(e*x+d)^(3/2)/b-2/b^2*a*e*(e*x+d)^(1/2)+2/b*(e*x+d)^(1/2)*d+2/b^2/((a*e-b*d)*b)^(1/2)*arctan((e*x+d)^(1/2)
/((a*e-b*d)*b)^(1/2)*b)*a^2*e^2-4/b/((a*e-b*d)*b)^(1/2)*arctan((e*x+d)^(1/2)/((a*e-b*d)*b)^(1/2)*b)*a*d*e+2/((
a*e-b*d)*b)^(1/2)*arctan((e*x+d)^(1/2)/((a*e-b*d)*b)^(1/2)*b)*d^2

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)*(e*x+d)^(3/2)/(b^2*x^2+2*a*b*x+a^2),x, algorithm="maxima")

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(a*e-b*d>0)', see `assume?` for
more details)Is a*e-b*d positive or negative?

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mupad [B]  time = 0.07, size = 93, normalized size = 1.08 \begin {gather*} \frac {2\,{\left (d+e\,x\right )}^{3/2}}{3\,b}-\frac {2\,\left (a\,e-b\,d\right )\,\sqrt {d+e\,x}}{b^2}+\frac {2\,\mathrm {atan}\left (\frac {\sqrt {b}\,{\left (a\,e-b\,d\right )}^{3/2}\,\sqrt {d+e\,x}}{a^2\,e^2-2\,a\,b\,d\,e+b^2\,d^2}\right )\,{\left (a\,e-b\,d\right )}^{3/2}}{b^{5/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((a + b*x)*(d + e*x)^(3/2))/(a^2 + b^2*x^2 + 2*a*b*x),x)

[Out]

(2*(d + e*x)^(3/2))/(3*b) - (2*(a*e - b*d)*(d + e*x)^(1/2))/b^2 + (2*atan((b^(1/2)*(a*e - b*d)^(3/2)*(d + e*x)
^(1/2))/(a^2*e^2 + b^2*d^2 - 2*a*b*d*e))*(a*e - b*d)^(3/2))/b^(5/2)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)*(e*x+d)**(3/2)/(b**2*x**2+2*a*b*x+a**2),x)

[Out]

Timed out

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